a: \(=\dfrac{5}{4}\cdot\dfrac{2}{5}\cdot\dfrac{17}{21}\cdot\dfrac{7}{34}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{12}\)

b: =>0,5^x(0,5+1)=1,5

=>0,5^x=1

=>x=0

c: =>x*0,05-0,25*x=-1,2

=>-0,3*x=-1,2

=>x=4

d: =>x-1=1 hoặc x-1=-1

=>x=0 hoặc x=2

e: =5+4=9

Giải:

\(\left(x.0,5-\dfrac{3}{7}\right):\dfrac{1}{2}=1\dfrac{1}{7}\)

\(\Leftrightarrow\left(x.0,5-\dfrac{3}{7}\right).2=\dfrac{8}{7}\)

\(\Leftrightarrow x-\dfrac{6}{7}=\dfrac{8}{7}\)

\(\Leftrightarrow x=\dfrac{8}{7}+\dfrac{6}{7}\)

\(\Leftrightarrow x=\dfrac{14}{7}=2\)

Vậy ...

\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)

c, ĐK: \(x>-7\)

\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)

Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)

d, ĐK: \(x>\dfrac{1}{2}\)

\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)

Kết hợp với ĐKXĐ, ta được: \(x\ge8\)

a: \(\Leftrightarrow x^2=900\)

=>x=30 hoặc x=-30

b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)

=>0,1x=2/3:50/3=2/3x3/50=1/25

=>1/10x=1/25

hay x=1/25:1/10=10/25=2/5

d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)

=>x=12/5 hoặc x=-12/5

\(a,\left(-\frac{1}{2}-0,6\right).x=\left(-3\right)^2+\frac{1}{2}\)

=> \(-\frac{11}{10}.x=9+\frac{1}{2}\)

=> \(-\frac{11}{10}.x=\frac{19}{2}\)

=> \(x=\frac{19}{2}:-\frac{11}{10}=-\frac{95}{11}\)

Vậy \(x\in\left\{-\frac{95}{11}\right\}\)

\(b,\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-2\right)x=0,5\)

=> \(\frac{1}{3}-\frac{1}{3}x=0,5\)

=> \(\frac{1}{3}x=\frac{1}{3}-0,5=-\frac{1}{6}\)

=> \(x=-\frac{1}{6}:\frac{1}{3}=-\frac{1}{2}\)

Vậy \(x\in\left\{-\frac{1}{2}\right\}\)

\(c,\left|x\right|+0,325=2\)

=> \(\left|x\right|=2-0,325=1,675=\frac{67}{40}\)

=> \(\left[{}\begin{matrix}x=\frac{67}{40}\\x=-\frac{67}{40}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{67}{40};-\frac{67}{40}\right\}\)

\(d,\left|x+0,25\right|=6,25\)

=> \(\left[{}\begin{matrix}x+0,25=6,25\\x+0,25=-6,25\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=6,25-0,25=6\\x=-6,25-0,25-6.5=-\frac{13}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{6;-\frac{13}{2}\right\}\)

a) \(\left(-\frac{1}{2}-0,6\right).x=\left(-3\right)^2+\frac{1}{2}\)

=> \(-\frac{11}{10}.x=9+\frac{1}{2}\)

=> \(-\frac{11}{10}.x=\frac{19}{2}\)

=> \(x=\frac{19}{2}:\left(-\frac{11}{10}\right)\)

=> \(x=-\frac{95}{11}\)

Vậy \(x=-\frac{95}{11}\).

b) \(\frac{1}{3}-\left(-\frac{1}{6}\right).\left(-2\right).x=0,5\)

=> \(\frac{1}{3}-\frac{1}{3}.x=\frac{1}{2}\)

=> \(\frac{1}{3}.x=\frac{1}{3}-\frac{1}{2}\)

=> \(\frac{1}{3}.x=-\frac{1}{6}\)

=> \(x=\left(-\frac{1}{6}\right):\frac{1}{3}\)

=> \(x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}\).

c) |x| + 0, 325 = 2

|x| = 2 - 0, 325

|x| = 1, 675

|x| = \(\frac{67}{40}\)

=> \(\left\{{}\begin{matrix}x=\frac{67}{40}\\x=-\frac{67}{40}\end{matrix}\right.\)

Vậy x ∈ \(\left\{\frac{67}{40};-\frac{67}{40}\right\}\).

d) |x + 0, 25| = 6, 25

TH1: x + 0, 25 = 6, 25

x = 6, 25 - 0, 25

x = 6

TH2: x + 0, 25 = -6, 25

x = (-6, 25) - 0, 25

x = -6, 5

Vậy x ∈ \(\left\{6;-6,5\right\}\).

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